# For the limit $\lim_{x\to -1}\frac{\sqrt{x+5}-1}{x+2}$, find $\delta$ that works for given $\epsilon$

The question is:

I've tried it many times and no matter what I do I always get a negative delta.

• How do you get a negative $\delta$ ? Nov 16, 2015 at 17:40

This will seem a little long, but I'm going to try to break this down as finely as possible, to make the process clearer.

One way to approach this is to go backwards first, starting from the thing you want to prove. Provided that you use invertible transformations, you should be able to reverse the analysis for your actual demonstration.

We begin with the desired condition

$$\left| \frac{\sqrt{x+5}-2}{x+1} - \frac{1}{4} \right| < \frac{1}{100}$$

Both the numerator and the denominator go to $0$ at $x = -1$, but the numerator has that square root in it. We might therefore get some mileage out of multiplying this fraction by $1$, in the form of $\sqrt{x+5}+2$ divided by itself, in order to produce a difference of squares and eliminate that square root:

$$\left| \frac{\sqrt{x+5}-2}{x+1} \cdot \frac{\sqrt{x+5}+2}{\sqrt{x+5}+2} - \frac{1}{4} \right| < \frac{1}{100}$$ $$\left| \frac{x+5-4}{(x+1)(\sqrt{x+5}+2)} - \frac{1}{4} \right| < \frac{1}{100}$$ $$\left| \frac{x+1}{(x+1)\left(\sqrt{x+5}+2\right)} - \frac{1}{4} \right| < \frac{1}{100}$$

Now, provided that we assert that $x \not= -1$, we can cancel out the $x+1$ terms, and obtain

$$\left| \frac{1}{\sqrt{x+5}+2} - \frac{1}{4} \right| < \frac{1}{100}$$

It may seem as though we've merely exchanged one square root for another, and not gained anything in the process. However, in this case, the $2$ is added to the square root, so the denominator does not vanish at $x = -1$. Let's gather the two fractions together at this point:

$$\left| \frac{2-\sqrt{x+5}}{4\left(\sqrt{x+5}+2\right)} \right| < \frac{1}{100}$$

Let us multiply again by $2+\sqrt{x+5}$ divided by itself, and we get

$$\left| \frac{-x-1}{4\left(\sqrt{x+5}+2\right)^2} \right| < \frac{1}{100}$$ $$\left| \frac{-x-1}{4\left(x+9+4\sqrt{x+5}\right)} \right| < \frac{1}{100}$$

At this point, imagine that we set $x = -1+\Delta x$, where $|\Delta x| < \delta$. We then have

$$\left| \frac{-\Delta x}{4\left(\Delta x+8+4\sqrt{\Delta x+4}\right)} \right| < \frac{1}{100}$$

We are now prepared to reverse our analysis to produce the demonstration. We observe that for small $\Delta x$, the denominator is surely greater than $1$. Therefore, let us set $\delta = 1/100$, and then

$$|\Delta x| < \frac{1}{100}$$

Multiplying the argument of the absolute value by $-1$ doesn't change it:

$$|-\Delta x| < \frac{1}{100}$$

And finally, if we divide something by an amount greater than $1$, we must surely reduce its absolute value:

$$\left| \frac{-\Delta x}{4\left(\Delta x+8+4\sqrt{\Delta x+4}\right)} \right| < \frac{1}{100}$$

We have now reached a point where we can "go backwards" and obtain the desired condition.

I repost it because I encountered a fatal error on the first step

$\left|\frac{\sqrt{x+5}-2}{x+1}-\frac{1}{4}\right|$

$=\left|\frac{4\sqrt{x+5}-x-9}{4(x+1)}\right|$

$=\left|\frac{16(x+5)-(x+9)^2}{4(x+1)(\sqrt{x+5}+x+9)}\right|$

$=\left|\frac{-x^2-2x-1}{4(x+1)(\sqrt{x+5}+x+9)}\right|$

$=\left|\frac{x+1}{4(\sqrt{x+5}+x+9)}\right|$

$<\left|\frac{x+1}{4}\right|$, as $x \to -1$

$<\frac{1}{100}$

.......Something like that